Equation of a plane, given two points and a perpendicular plane

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The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.




What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?




vectors analytic-geometry




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    Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
    – Hagen von Eitzen
    Sep 18 '16 at 14:42










  • $(7,9,4)$ is a vector, not a plane.
    – anderstood
    Sep 18 '16 at 14:42






  • 1




    Also, the cross product is $(51,-49,21)$. The 49 is negative.
    – Nick
    Sep 18 '16 at 14:43










  • You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
    – G Cab
    Sep 18 '16 at 14:44











  • Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
    – Arghya
    Sep 18 '16 at 16:18














up vote
1
down vote

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The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.




What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?




vectors analytic-geometry




share|cite|improve this question


















  • 1




    Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
    – Hagen von Eitzen
    Sep 18 '16 at 14:42










  • $(7,9,4)$ is a vector, not a plane.
    – anderstood
    Sep 18 '16 at 14:42






  • 1




    Also, the cross product is $(51,-49,21)$. The 49 is negative.
    – Nick
    Sep 18 '16 at 14:43










  • You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
    – G Cab
    Sep 18 '16 at 14:44











  • Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
    – Arghya
    Sep 18 '16 at 16:18












up vote
1
down vote

favorite









up vote
1
down vote

favorite












The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.




What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?




vectors analytic-geometry




share|cite|improve this question















The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.




What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?





vectors analytic-geometry





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 18 '16 at 15:03

























asked Sep 18 '16 at 14:38









Arghya

134




134







  • 1




    Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
    – Hagen von Eitzen
    Sep 18 '16 at 14:42










  • $(7,9,4)$ is a vector, not a plane.
    – anderstood
    Sep 18 '16 at 14:42






  • 1




    Also, the cross product is $(51,-49,21)$. The 49 is negative.
    – Nick
    Sep 18 '16 at 14:43










  • You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
    – G Cab
    Sep 18 '16 at 14:44











  • Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
    – Arghya
    Sep 18 '16 at 16:18












  • 1




    Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
    – Hagen von Eitzen
    Sep 18 '16 at 14:42










  • $(7,9,4)$ is a vector, not a plane.
    – anderstood
    Sep 18 '16 at 14:42






  • 1




    Also, the cross product is $(51,-49,21)$. The 49 is negative.
    – Nick
    Sep 18 '16 at 14:43










  • You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
    – G Cab
    Sep 18 '16 at 14:44











  • Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
    – Arghya
    Sep 18 '16 at 16:18







1




1




Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42




Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42












$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42




$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42




1




1




Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43




Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43












You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44





You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44













Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18




Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18










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make the ansatz the plane has the equation
$$ax+by+cz+d=0$$ then we have
$$3a+4b+c+d=0$$(1)
$$3a+b-6c+d=0$$ (2)
and
$$[a,b,c]cdot[7,9,4]=0$$ (3)
can you proceed?






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    make the ansatz the plane has the equation
    $$ax+by+cz+d=0$$ then we have
    $$3a+4b+c+d=0$$(1)
    $$3a+b-6c+d=0$$ (2)
    and
    $$[a,b,c]cdot[7,9,4]=0$$ (3)
    can you proceed?






    share|cite|improve this answer
























      up vote
      0
      down vote













      make the ansatz the plane has the equation
      $$ax+by+cz+d=0$$ then we have
      $$3a+4b+c+d=0$$(1)
      $$3a+b-6c+d=0$$ (2)
      and
      $$[a,b,c]cdot[7,9,4]=0$$ (3)
      can you proceed?






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        make the ansatz the plane has the equation
        $$ax+by+cz+d=0$$ then we have
        $$3a+4b+c+d=0$$(1)
        $$3a+b-6c+d=0$$ (2)
        and
        $$[a,b,c]cdot[7,9,4]=0$$ (3)
        can you proceed?






        share|cite|improve this answer












        make the ansatz the plane has the equation
        $$ax+by+cz+d=0$$ then we have
        $$3a+4b+c+d=0$$(1)
        $$3a+b-6c+d=0$$ (2)
        and
        $$[a,b,c]cdot[7,9,4]=0$$ (3)
        can you proceed?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 18 '16 at 15:07









        Dr. Sonnhard Graubner

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